Question

You are conducting a study to see if the proportion of teenagers between 15 and 19 that drink alconol is significantly different from 0.37. You use a significance level of α=0.002α=0.002.

      H0:p=0.37H0:p=0.37
      H1:p≠0.37H1:p≠0.37

You obtain a sample of size n=570n=570 in which there are 199 successes.

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This p-value leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the proportion of teenagers between 15 and 19 that drink alconol is different from 0.37.
• There is not sufficient evidence to warrant rejection of the claim that the proportion of teenagers between 15 and 19 that drink alconol is different from 0.37.
• The sample data support the claim that the proportion of teenagers between 15 and 19 that drink alconol is different from 0.37.
• There is not sufficient sample evidence to support the claim that the proportion of teenagers between 15 and 19 that drink alconol is different from 0.37.

Answer 1

Solution:

Given:

Level of significance =  α = 0.002

We have to test if the proportion of teenagers between 15 and 19 that drink alcohol is significantly different from 0.37.

Hypothesis of the study are:

H_o : p = 0.37 Vs H_a : p  $\neq$ 0.37

Alternative hypothesis H_a is the claim.

Sample size = n = 570

x = 199

Find p-value:

to find p-value find z test statistic.

z test stat for proportion

-P px(1-P)

where

199 = 0.3491

thus

-P px(1-P)

0.3491 -0.37 0.37x(1-0.37) 570

-0.0209 0.37x0.63 570

-0.0209 0.000408947

-0.0209 10.02022245

2 = -1.03

P-value for two tailed test when z is negative:

P-value = 2 X P( Z< z value)

P-value = 2 X P( Z< -1.03)

Look in z table for z = -1.0 and 0.03 and find corresponding area.

| .00 .01 .02 | .03 -3.4 .0003 .0003 0003 .0003 -3.3 .0005 .0005 .0005 1004 -3.2 .0007 ,0007 ,0006 .06 -3.1 .0010 ,0009 10000 ,000 -3.0 .0013 0013 .0013 中12 -2.9 .0019 .0018 .0018 017 -2.8 .0026 .0025 .0024 .023 -2.7 .0035 .0034 .0033 1.32 -2.6 .0047 .0045 .0044 .43 -2.5 .0062 .0060 .0059 057 -2.4 .0082 0080 .0078 .0075 -23 010701040102 1099 -2.2 .0139 .0136 .0132 129 -2.1 .1790174 010 1166 -2.0 .0228 .0222 ,0217 D212 -1.9 .0287 .0281 .0274 D268 1-1.8 0359 .0351 .0344 336 -1.7 .0446 0436 .0427 0418 -1.6 .0548 ,0537 0526 1.1516 -1.5 .0668 .0655 1.0643 0530 -1.4 .0808 .0793 .0778 064 -13 0968 ,0951 0934 ,0918 -1.2 .1151 .1131 .1112 .1493 -1.1 .137 .1335 1314 1292 -1.0 -1537-A2 1539-1515 .04 .0003 ,0004 .0006 ,0008 0012 10016 .0023 .0031 .0041 .0055 .0073 006 .0125 .0152 .0207 .0262 .0329 10409 .0505 10618 .0749 .0901 175 1271 1492 .05 .06 .07 .08 .09 .0003 .000 .0003 .0003 .0002 0004 0004 ,0004 ,0004 0003 .0006 .0006 .0005 .0005 .0005 10008 .0008 .0008 .0007 ,0007 .00 11 .0011.0011 .00100010 0016 .0015 ,0015 .0014 2014 .0022 .0021 .0021 .0020 .0019 .0030 .0029 .0028 .0027 .0026 .0040 .0039 .0038 .0037 0036 .0054 .0052 .0051 .0049 0048 0071 .0069 0068 .0066 .0064 0094 0091 0089 .0087 .0084 .01220119 0116 01130110 .0158 .014 0150 0146 0143 .0202 .0197 .0192 .0188 .0183 .0256 .0250 .0244 .0239 .0233 ,0322.0314.030703010294 .0401 .0392 0384 .0375 0367 .0495 .0485 .0475 .0465 .0455 10606 0594 ,0582 .071 ,0559 .0735 .0721 .0708 .0699 .0681 ,0885 ,0869 ,0853 ,0838 0 823 1056 1038 .1020 .1003 0985 .1251 1230 .1210 .1190 1170 .1469.146 .1423 .1401 1379

P( Z < -1.03 ) = 0.1515

Thus

P-value = 2 X P( Z< -1.03)

P-value = 2 X 0.1515

P-value = 0.3030

The p-value = 0.3030 is greater than α = 0.002.

This p-value leads to a decision to fail to reject the null

As such, the final conclusion is that:  There is not sufficient sample evidence to support the claim that the proportion of teenagers between 15 and 19 that drink alcohol is different from 0.37.